Bundor-PHY150-M2-KinematicsLabReport

V=0.667 m/s Question 3: What does the slope of the line during each time interval represent? The slope of the line on a position-time graph represents the velocity of the object being tracked. Question 4: From time t = 35 s until t = 45 s, the train is located at the same position. What is slope of the line while the train is stationary?  The slope of the line while the train is stationary (from t=35t=35 s to t=45t=45 s) is zero. This is because the position of the train remains constant during this time interval, meaning there is no change in position with respect to time. Question 5: Calculate the average speed of the train as it moves from position x = 70m to x = 55m. What does the sign of the average velocity during this time interval represent? Total Time=t2−t1=50s−40s=10s Now, plug these values into the formula for average speed: Average Speed=15 m10 s=1.5 m/s The sign of the average velocity during this time interval represents the direction of motion. Since the velocity is positive (1.5 m/s), it indicates that the train is moving towards the positive direction, which means it's moving to the left on the position- time graph. Question 6: What is the displacement of the train from time t = 0s until t = 50s? Δx=x2−x1 Δx=55 m−0 Δx=55 m So, the displacement of the train from t=0s until t=50s is 55 meters. Question 7: What is the total distance traveled by the train from time t = 0s until t = 50s? Total Distance= ∣ 20−0 ∣ + ∣ 40−20 ∣ + ∣ 50−40 ∣ + ∣ 55−50 ∣ + ∣ 60−55 ∣ + ∣ 70−60 ∣ + ∣ 70−70 ∣ + ∣ 70− 70 ∣ + ∣ 55−70 ∣ Total Distance=20+20+10+5+5+10+0+0+15 Total Distance=85 meters So, the total distance traveled by the train from t=0t=0 s until t=50t=50 s is 8585 meters. Question 8. What is the slope of the line during the time interval t = 45 to t = 50?

Activity 3. Data Table 1. Time (s) Average Time (s) Average Time 2 (s 2 ) Distance (m) Trial 1 =0 0 0 0 Trial 2 =0 Trial 3 =0 Trial 1 =1.33 1.49 2.2201 0.1 Trial 2 =1.42 Trial 3 =1.73 Trial 1 =1.54 1.48 2.2002 0.2 Trial 2 =2.03 Trial 3 =0.88 Trial 1 =1.38 1.10 1.2031 0.3 Trial 2 =0.89 Trial 3 =1.02 Trial 1 =1.46 1.77 3.1447 0.4 Trial 2 =2.47 Trial 3 =1.39 Trial 1 =0.79 1.56 2.4440 0.5 Trial 2 =1.53 Trial 3 =2.37 Trial 1 =2.01 1.86 3.4596 0.6 Trial 2 =2.33 Trial 3 =1.24 Trial 1 =1.09 1.40 1.9693 0.7 Trial 2 =1.67 Trial 3 =1.45 Trial 1 =1.36 1.35 1.8225 0.8 Trial 2 =0.99 Trial 3 =1.70 *Note that 0.10 m = 10 cm Insert your graphs of Distance vs Time (m) and Distance vs Time Squared here: Questions for Activity 3 Question 1: What is the shape of the graph when displacement is graphed vs. time? A. When I graph displacement against time using the provided data, I observe that the shape of the graph appears like a scattered plot instead of a smooth continuous curve. This occurs because the data points